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    Q. Find the set of values of x for which:

    4x^2 - 3x - 1 < 0 and 4(x+20) < 15 - (x + 7)

    Cant seem to factorise the quadratic
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    (Original post by Modesty)
    Q. Find the set of values of x for which:

    4x^2 - 3x - 1 < 0 and 4(x+20) < 15 - (x + 7)

    Cant seem to factorise the quadratic
    In usual fashion you want to find two numbers that add to make -3 and multiply to make -4.
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    (4X+1)(X-1)
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    (Original post by 13 1 20 8 42)
    In usual fashion you want to find two numbers that add to make -3 and multiply to make -4.
    Whoh.
    The way i factorise is different i will put (4x ) (4x )
    Find two numbers multiplyto give -3 add to give -1 i'm confused as to which two numbers??
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    (Original post by dons178)
    (4X+1)(X-1)
    I factorise differently , i put 4x in the beginning of both brackets but stuck
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    (Original post by Modesty)
    Whoh.
    The way i factorise is different i will put (4x ) (4x )
    Find two numbers multiplyto give -3 add to give -1 i'm confused as to which two numbers??
    You don't put 4x at the beginning of both. Think about what happens when you expand it; you'll get 16x^2.

    You either do (4x )(x ) or (2x )(2x ).
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    (Original post by Modesty)
    Whoh.
    The way i factorise is different i will put (4x ) (4x )
    Find two numbers multiplyto give -3 add to give -1 i'm confused as to which two numbers??
    Well assuming you mean like (4x + ?)(4x + ?) that wouldn't work here..
    multiply to make -4. You generally just want to exhaust the factors, positive and negative of the number AC in Ax^2 + Bx + C, until you find the two that also add to make B. Then you write out the x-term as the sum of two x terms whose coefficients are the two numbers and this allows you to factorize. Eventually you'll probably be able to factorize expressions like this very quickly just by experience/recognition
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    (Original post by ubisoft)
    You don't put 4x at the beginning of both. Think about what happens when you expand it; you'll get 16x^2.
    That's how i factorise quadratics when coefficient is greater than 1.
    Why is it with this question i can't?
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    (Original post by Modesty)
    That's how i factorise quadratics when coefficient is greater than 1.
    Why is it with this question i can't?
    Well it is wrong, don't do that. Expand it out and you will see it's not the same.
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    (Original post by 13 1 20 8 42)
    Well assuming you mean like (4x + ?)(4x + ?) that wouldn't work here..
    multiply to make -4. You generally just want to exhaust the factors, positive and negative of the number AC in Ax^2 + Bx + C, until you find the two that also add to make B. Then you write out the x-term as the sum of two x terms whose coefficients are the two numbers and this allows you to factorize. Eventually you'll probably be able to factorize expressions like this very quickly just by experience/recognition
    So answer is (4x-4) (x+1)
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    (Original post by Modesty)
    So answer is (4x-4) (x+1)
    No. Note that 4x^2 - 3x - 1 = 4x^2 - 4x + x - 1. can you see how to factorize this?
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    (Original post by 13 1 20 8 42)
    No. Note that 4x^2 - 3x - 1 = 4x^2 - 4x + x - 1. can you see how to factorize this?
    Ahhhh i'm so stupid. I just remembered these are Trinomials.

    Mutliply 4 * -1 = -4
    Find two numbers which mutliply to give -4 and add to give 3
    So (4x+4) (4x-1)

    Now divide first bracket by 4 which gives x+1

    Answer is (x+1) (4x-1)
 
 
 
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